One end of a long rod 3 cm in diameter is inserted into a furnace with the outer end projecting into the outside air. Once the steady state is reached the temperature of the rod is measured at two points, 15 cm apart and found to be 140°C and 100°C, when the atmospheric air is at 30°C with convection coefficient of 20 W/m^2.K.20W/m 2 .K. Calculate the thermal conductivity of the rod material.
Answers
Answer:
Given : One end of the long rod inserted into a furnace ;
d = 3 cm = 0.03 m, L = 15 cm = 0.15 m,
T_0 = 140°C, T_∞ = 30°C,T
0
=140°C,T
∞
=30°C,
T_L = 100°C, h = 20 W/m^2.KT
L
=100°C,h=20W/m
2
.K
To find : The thermal conductivity of the rod material.
Assumptions :
(i) Steady state conditions.
(ii) One dimensional conduction along the rod.
(iii) Constant properties.
(iv) No internal heat generation.
(v) Infinite long fin.
Analysis : For infinite long fin, the temperature distribution is given by eqn. (5.13)
\frac{T(x) – T_∞}{T_0 – T_∞}= e^{–mx}
T
0
–T
∞
T(x)–T
∞
=e
–mx
The starting point,
at x = 0, T_0 = 140°Cx=0,T
0
=140°C
and at x = L = 0.15 m, T_L = 100°Cx=L=0.15m,T
L
=100°C
Thus \frac{100 – 30}{140 – 30} = e^{–m × 0.15}
140–30
100–30
=e
–m×0.15
It gives m = 3.013
We have m =\sqrt{\frac{ hP}{k A_c}}m=
kA
c
hP
Thus \sqrt{\frac{20 × (\pi × 0.03)}{k ×{(\pi /4) × (0.03)^2}}} = 3.013
k×(π/4)×(0.03)
2
20×(π×0.03)
=3.013
It gives k = 293.74 W/m.K.
Step-by-step explanation:
Given : One end of the long rod inserted into a furnace ;
d = 3 cm = 0.03 m, L = 15 cm = 0.15 m,
T_0 = 140°C, T_∞ = 30°C,T
0
=140°C,T
∞
=30°C,
T_L = 100°C, h = 20 W/m^2.KT
L
=100°C,h=20W/m
2
.K
To find : The thermal conductivity of the rod material.
Assumptions :
(i) Steady state conditions.
(ii) One dimensional conduction along the rod.
(iii) Constant properties.
(iv) No internal heat generation.
(v) Infinite long fin.
Analysis : For infinite long fin, the temperature distribution is given by eqn. (5.13)
\frac{T(x) – T_∞}{T_0 – T_∞}= e^{–mx}
T
0
–T
∞
T(x)–T
∞
=e
–mx
The starting point,
at x = 0, T_0 = 140°Cx=0,T
0
=140°C
and at x = L = 0.15 m, T_L = 100°Cx=L=0.15m,T
L
=100°C
Thus \frac{100 – 30}{140 – 30} = e^{–m × 0.15}
140–30
100–30
=e
–m×0.15
It gives m = 3.013
We have m =\sqrt{\frac{ hP}{k A_c}}m=
kA
c
hP
Thus \sqrt{\frac{20 × (\pi × 0.03)}{k ×{(\pi /4) × (0.03)^2}}} = 3.013
k×(π/4)×(0.03)
2
20×(π×0.03)
=3.013
It gives k = 293.74 W/m.K.